Solve {ccc}{k}&{-2}&{1}{2}&{0}&{k}{3}&{-2}&{2}

$$\left| \begin{array} { c c c } { k } & { - 2 } & { 1 } \\ { 2 } & { 0 } & { k } \\ { 3 } & { - 2 } & { 2 } \end{array} \right|$$

Evaluate

$$2\left(k-2\right)\left(k-1\right)$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}k&-2&1\\2&0&k\\3&-2&2\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}k&-2&1&k&-2\\2&0&k&2&0\\3&-2&2&3&-2\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$-2k\times 3+2\left(-2\right)=-6k-4$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$-2kk+2\times 2\left(-2\right)=-2k^{2}-8$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$-6k-4-\left(-2k^{2}-8\right)$$

Subtract -2k^{2}-8 from -6k-4.

$$2\left(k-2\right)\left(k-1\right)$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}k&-2&1\\2&0&k\\3&-2&2\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$kdet(\left(\begin{matrix}0&k\\-2&2\end{matrix}\right))-\left(-2det(\left(\begin{matrix}2&k\\3&2\end{matrix}\right))\right)+det(\left(\begin{matrix}2&0\\3&-2\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$k\left(-\left(-2k\right)\right)-\left(-2\left(2\times 2-3k\right)\right)+2\left(-2\right)$$

Simplify.

$$k\times 2k-\left(-2\left(4-3k\right)\right)-4$$

Add the terms to obtain the final result.

$$2\left(k-2\right)\left(k-1\right)$$

Integrate w.r.t. k

$$\frac{2k^{3}}{3}-3k^{2}+4k+С$$